Question: A few families took a trip to an amusement park together. Tickets cost $$5.00$ each for adults and $$4.50$ each for kids, and the group paid $$47.00$ in total. There were $2$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Answer: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${5x+4.5y = 47}$ ${x = y-2}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-2}$ for $x$ in the first equation. ${5}{(y-2)}{+ 4.5y = 47}$ Simplify and solve for $y$ $ 5y-10 + 4.5y = 47 $ $ 9.5y-10 = 47 $ $ 9.5y = 57 $ $ y = \dfrac{57}{9.5} $ ${y = 6}$ Now that you know ${y = 6}$ , plug it back into ${x = y-2}$ to find $x$ ${x = }{(6)}{ - 2}$ ${x = 4}$ You can also plug ${y = 6}$ into ${5x+4.5y = 47}$ and get the same answer for $x$ ${5x + 4.5}{(6)}{= 47}$ ${x = 4}$ There were $4$ adults and $6$ kids.